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I'm in the process of learning how transistors works, which starts with understanding how doping is used to create n-type and p-type semiconductor materials.

All the resources I've read sort of explain this the same way, and I'm missing something. P-type semiconductors have extra holes and are predisposed to accept electrons, whereas n-type semiconductors have extra free electrons and are predisposed to donate them. This is the fundamental principle of how transistors work, as I understand it.

But every resource emphasizes that in spite of this both n-type and p-type semiconductors are electrically neutral, which is where I'm lost. If one has extra electrons, and one is missing electrons, how are they electrically neutral and not charged? I seem to have a block about this or something, I just don't get it.

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  • \$\begingroup\$ It just means that the overall charge content of the wafer of either P and N material remains the same(equal to initial charge),as the breakup of immobile ions result in both the pairs which overall sum to zero,and each time a electron enter from the cathode one exits from the anode,hence overall neutrality is maintained. \$\endgroup\$
    – MaMba
    May 29 '15 at 10:39
  • \$\begingroup\$ n-type does not have any extra electrons. It has the same number of negative electrons as positive static atomic nuclei. What makes it "n-type" is that some electrons are not bound to static nuclei and thus can move freely (and thus, act as charge carriers and conduct current). The same logic applies to p-type as well. \$\endgroup\$
    – akhmed
    Aug 19 '16 at 0:34
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Take silicon as an example. Silicon has four valence electrons, and silicon atoms in a crystal lattice form four bonds with neighbouring atoms.

Transistors, and other semiconductors, are made of silicon crystal with small amounts of dopants added. These dopants change the electrical properties because of the way they interact with the crystal lattice. Phosphorous, for example, has 5 valence electrons. It's still electrically neutral (number of protons = number of electrons) but since the silicon crystal structure only requires 4 bonds per atom, there is an 'extra' electron that isn't really participating in the crystal structure. With a bit of extra energy, that electron will go into the conduction band and freely roam around the crystal lattice. This corresponds to an n-type semiconductor.

There is a similar process for p-type semiconductors - boron, for example, only has 3 valence electrons.

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Good answer can be found here, taken from a physical standpoint.

The terms n- and p-type doped do only refer to the majority charge carriers. Each positive or negative charge carrier belongs to a fixed negative or positive charged dopant.

p and n type materials are NOT positively and negatively charged.

An n-type material by itself has mainly negative charge carriers (electrons) which are able to move freely, but it is still neutral because the fixed donor atoms, having donated electrons, are positive.

Similarly p-type material by itself has mainly positive charge carrier (holes) which are able to move relatively freely, but it is still neutral because the fixed acceptor atoms, having accepted electrons, are negative.

https://physics.stackexchange.com/questions/81488/how-can-doped-semiconductor-be-neutral

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The semiconductor has both free charge (electrons and holes) and immobile charge (lower band electrons, nuclear protons, and ionized donors and acceptors).

When a donor (for example) is ionized, it creates a free electron, but also it creates a positively ionized donor atom. The charge on the free electron and the ionized donor are equal and opposite. So as long as the electron doesn't go anywhere, the net charge remains zero.

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They are not always electrically neutral.

An n-type semiconductor has an excess of 'free' electrons -- electrons that can move about freely in the semiconductor (very similar to electrons in a metal). These electrons are 'donated' by immobile donor impurities doped in to the semiconductor.

If you imagine starting from that state, then the result is still neutral. However since the electrons can move, they have a tendency to diffuse away from regions of high concentration. If you connect another material (e.g. p-type) to the n-type (forming a pn junction), electrons will diffuse from the high concentration region to the low-concentration region. This won't continue forever (unless you have a power source connected), because in leaving the n-type region, they leave a + charge behind. This creates a restoring electric field, and at some point this restoring field will balance the diffusion process and an equilibrium will be obtained. The specifics of this depend on the materials, the doping and temperature, as well as any external voltage applied between the 2 materials forming the p-n junction.

Since (starting from neutral), electrons (negative charge) have left the n-type region, it will become net positively charged, and the p-type negatively charged. In a similar way holes ('anti-electrons') from the p-type diffuse over to the n-type, further charging it positively.

A similar behaviour would occur if you connected a heavily doped n-type to a lightly-doped (in fact it occurs any time there is a concentration (or temperature) gradient).

The material as a whole isn't charged (just polarized), but if you connected it to another conductor (e.g. a wire), charge would move between the cloud of free electrons in the wire to the semiconductor, putting a net negative charge on it. Although it is small, it could in principle be detected by observing electrostatic forces. It cannot be measured by (e.g.) connecting a voltmeter to the semiconductor and the metal because charges would also flow into the leads of the voltmeter, exactly canceling and leaving no net voltage. If in fact there was a temperature difference, you would be able to measure a voltage -- this is the Seebeck (thermocouple) effect.

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Simple version:

When n-doping by adding phosphorus, we're actually adding a positive phosphorus ion, plus a mobile electron.

When p-doping by adding Boron, we're actually adding a negative boron ion, as well as a mobile "hole."

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